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प्रश्न
Differentiate the following w.r.t. x : `cot^-1((a^2 - 6x^2)/(5ax))`
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उत्तर
Let y = `cot^-1((a^2 - 6x^2)/(5ax))`
= `tan^-1((5ax)/(a^2 - 6x^2)) ...[∵ cot^-1 x = tan^-1(1/x)]`
= `tan^-1[(5(x/a))/(1 - 6(x/a)^2)]` ...[Dividing by a2]
= `tan[(3(x/a) + 2(x/a))/(1 - 3(x/a) xx 2(x/a))]`
= `tan^-1((3x)/a) + tan^-1((2x)/a)`
Differentiating w.r.t. x, we get
`"dy"/"dx"= "d"/"dx"[tan^-1((3x)/a) + tan^-1((2x)/a)]`
= `"d"/"dx"[tan^-1((3x)/a)] + "d"/"dx"[tan^-1((2x)/a)]`
= `1/(1+((3x)/a)^2)*d/dx((3x)/a)+1/(1+((2x)/a)^2)*d/dx((2x)/a)`
= `(1)/(1 + ((9x^2)/a^2)) xx (3)/a xx 1 + (1)/(1+((4x^2)/a^2)) xx (2)/a xx 1`
= `a^2/(a^2 + 9x^2) xx (3)/a + a^2/(a^2 + 4x^2) xx (2)/a`
= `(3a)/(a^2 + 9x^2) + (2a)/(a^2 + 4x^2)`
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